For what value of k, the function
`f(x) ={:{(2x+1", "x gt2),(" "k ", " x=2),(3x-1 ", "x lt2):},`
is continuous at x=2.

Answer

To determine the value of \( k \) for which the function

\[ f(x) = \begin{cases} 2x + 1 & \text{if } x > 2 \\ k & \text{if } x = 2 \\ 3x - 1 & \text{if } x < 2 \end{cases} \]

is continuous at \( x = 2 \), we need to ensure that the left-hand limit (LHL), right-hand limit (RHL), and the function value at that point are all equal.

### Step 1: Calculate the Left-Hand Limit (LHL) as \( x \) approaches 2 from the left.

The left-hand limit is given by the expression for \( f(x) \) when \( x < 2 \):

\[ \text{LHL} = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (3x - 1) \]

Substituting \( x = 2 \):

\[ \text{LHL} = 3(2) - 1 = 6 - 1 = 5 \]

### Step 2: Calculate the Right-Hand Limit (RHL) as \( x \) approaches 2 from the right.

The right-hand limit is given by the expression for \( f(x) \) when \( x > 2 \):

\[ \text{RHL} = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x + 1) \]

Substituting \( x = 2 \):

\[ \text{RHL} = 2(2) + 1 = 4 + 1 = 5 \]

### Step 3: Set the limits equal to the function value at \( x = 2 \).

For the function to be continuous at \( x = 2 \), we need:

\[ \text{LHL} = \text{RHL} = f(2) \]

From our calculations:

\[ \text{LHL} = 5, \quad \text{RHL} = 5 \]

Thus, we have:

\[ f(2) = k \]

Setting the limits equal to the function value:

\[ k = 5 \]

### Conclusion

The value of \( k \) for which the function is continuous at \( x = 2 \) is:

\[ \boxed{5} \]